I'm trying to solve the following exercise.
Let $\gamma: \mathbb{R} \rightarrow \mathbb{R}^{2}$ be a regular plane curve without self-intersections and consider the set $S \subset \mathbb{R}^{3}$, given by $S=\left\{(x, y, z) \in \mathbb{R}^{3}:(x, y) \in \gamma(\mathbb{R})\right\}$. Show that $S \subset \mathbb{R}^{3}$ is a regular surface that is homeomorphic to a plane.
I have managed to show that such a surface is indeed a regular surface, but am struggling to come up with a suitable homeomorphism, or see why it is a homeomorphism. The official solutions state that "it is clear that the surface is homeomorphic to a plane", but I cannot understand why this is. Can anyone help me understand this?
Best Answer
What you might be overlooking is that $\gamma$ is a parameterized curve, meaning that you can write $\gamma$ in the form $$\gamma(t)=(\gamma_1(t),\gamma_2(t)), \quad t \in \mathbb R $$ Once you've done that, then the parameters $t$, $z$ define the homeomorphism you want: $$S = \{(\gamma_1(t),\gamma_2(t),z) \mid (t,z) \in \mathbb R^2\} $$ and the function $f : \mathbb R^2 \to S$ defined by $f(t,z) =(\gamma_1(t),\gamma_2(t),z)$ is the desired homeomorphism